A straight line through the point $(1, 1)$ meets the $x$-axis at ‘$A$’ and the $y$-axis at ‘$B$’. The locus of the mid-point of $AB$ is
$2xy + x + y = 0$
$x + y - 2xy = 0$
$x + y + 2 = 0$
$x + y - 2 = 0$
Let a triangle be bounded by the lines $L _{1}: 2 x +5 y =10$; $L _{2}:-4 x +3 y =12$ and the line $L _{3}$, which passes through the point $P (2,3)$, intersect $L _{2}$ at $A$ and $L _{1}$ at $B$. If the point $P$ divides the line-segment $A B$, internally in the ratio $1: 3$, then the area of the triangle is equal to
Triangle formed by the lines $3x + y + 4 = 0$ , $3x + 4y -15 = 0$ and $24x -7y = 3$ is a/an
Given $A(1, 1)$ and $AB$ is any line through it cutting the $x-$ axis in $B$. If $AC$ is perpendicular to $AB$ and meets the $y-$ axis in $C$, then the equation of locus of mid- point $P$ of $BC$ is
If the point $\left(\alpha, \frac{7 \sqrt{3}}{3}\right)$ lies on the curve traced by the mid-points of the line segments of the lines $x$ $\cos \theta+ y \sin \theta=7, \theta \in\left(0, \frac{\pi}{2}\right)$ between the coordinates axes, then $\alpha$ is equal to